Answer:
the total workdone required to to lift the chain from its top end so that its bottom is 2 meters above the ground = 661 J
Step-by-step explanation:
Given that:
The mass density of the chain is [tex]\rho (x) = 2x( 4-x) \ kg/m[/tex]
It is pertinent and crucial to consider the determination of the work-done in lifting the chain from its front in that its bottom is 2 meters away from the ground.
Consider a cross section portion of the chain of length Δx that has to be lifted to a height [tex]x_k[/tex]
The required wok to be done for this work is [tex]W_k = \rho(x_k)g \delta x(x_k)[/tex]
combining the segments of the chain and taking the [tex]\delta x[/tex]→0
Integrally, the work-done can be illustrated as :
[tex]\int\limits^3_0 \rho {(x_k)} \, gxdx \ \ = \ \ \int\limits^3_0 (9.8) 2x^2 (4-x)dx \\\\= 19.6 \int\limits^3_0 (4x^2-x^3)dx\\\\\\= 19.6 [ \frac{4}{3}x^2- \frac {x^4}{4}]^3__0}}dx\\\\\\= 19.6 [ \frac{4}{3}(3)^2- \frac {3^4}{4}]}dx\\\\= 19.6 (36- \frac{81}{4})\\\\\\= 308.7 \ \ J[/tex]
Furthermore, there is need to lift the chain up to 2 meters . So, calculating the weight of the chain ; we have:
Weight = [tex]\int\limits^3_0 \rho {x} \, gdx[/tex]
[tex]= \int\limits^3_0 (9.8) {2x}(4-x) \, dx\\ \\\\= 19.6\int\limits^3_0 {(4x-x^2)} \, dx \\\\= 19.6 [ 2x^2 - \frac{x^3}{3}]^3_0\\\\= 19.6 [2(3)^2 - \frac{3^3}{3}]\\\\=19.6 [18-9]\\\\= 176.4 \ \ J[/tex]
Finally .the work-done is said to be equal to the potential energy
∴ W = mgh
W = (176.4)×2
W = 352.8 J
Total workdone = (308.7 + 352.8 ) J
Total workdone = 661 J
Thus, the total workdone required to to lift the chain from its top end so that its bottom is 2 meters above the ground = 661 J
Answer:
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that,
Length of chain
L = 3m
Linear mass density is
ρ(x) = 2x(4 — x) kg/m lies on the ground
When x = 0, is top of the chain
Work done to lift the chain from top end so that the bottom is 2m above the ground.
Considered the segment of the chain of length ∆x that will be lifted in the positive y direction (+j)m from the foot.
The work needed to lift this segment is given as
Work = mass density × ∆x × gravity
W = ρ × ∆x × g
g is acting downward = 9.8j
Summing over all segment of the chain and passing to the limit as ∆x→0.
Therefore, the total work done needed to full extend the chain is
W = ∫ ρ × ∆x × g x = 0 to 3
Since g is constant
∆x = xdx
Then,
W = g ∫ 2x(4—x)x dx. x= 0 to x = 3
W = 9.81 ∫ (8x² — 2x³)x dx
W = 9.81 ( 8x³/3 — 2x⁴/4)
W=9.81(8x³/3— ½x⁴) from x=0 to x=3
W = 9.81[8(3)³/3 — ½(3⁴)] — 0
W = 9.81 × (72 —40.5)
W = 9.81 × 31.5
Work done= 309.015 J
Lifting the entire chain requires to light the weight
Weight = ∫ρgdx
Weight = g ∫ρ dx. From x=0 to x=3
Weight = g ∫ 2x(4-x) dx
Weight = 9.81 ∫(8x-2x²)dx
Weight = 9.81 [ 8x²/2 - 2x³/3]
Weight = 9.81[4x²-⅔x³] x=0 to x=3
Weight = 9.81[4(3²) — ⅔(3³)]
Weight = 9.81(36—18)
Weight = 9.81 × 18
Weight = 176.58N
Now, this weight is lifted to a height of 2m, then using potential energy formula, we have
P.E = Work = mgh = Weight ×height
Work = W×h = 176.58 × 2
Work = 353.16 J
Then, total workdone is
W = 353.16 + 309.015
W = 662.18 J
The required Work done required to lift the chain from top so that it's bottom is 2m from the ground is 662.18J
